Any photon that comes out of a polarisation filter, splitter or other polarising device will be in an eigenstate of that basis, i.e. it will have a collapsed wave function. That does not have to be true for the photon/wave function going into the polariser.

For example, a circularly polarized photon contains both |H> and |V> components. It will have a 50% chance of passing a |H><H|filter and coming out as |H> eigenstate.

(1) I wonder: If we have a pair of entangled photons (entangled as HH+VV), we send photon #2 to a polarising rotator orientated at 22.5 degrees. Prior to photon #2 reaching the subsequent PBS (in the H/V basis) - so all it has done is gone through the polarising rotator - we measure photon #1, also with a polarising rotator orientated at 22.5 degrees. Photon #1 also reaches the PBS and we detect the photon as H polarised.

Because the input of photon #2 into the polarising rotator was undefined at the time (photon #1 hasn't been detected as H), we get an undefined output. Now that photon #1 has acquired H polarisation, does the output at the polarising rotator that photon #2 went into now become: input H polarisation -> output 45 degree polarisation?

(2) In another case, if we had two photons entangled as HH+VV, and we send both through quarter wave plates, then only one goes onto PBS in H/V, do we write the outcome as H(output) + V(output) OR HH + VV, where (output) is the output polarisation of the 2nd photon not yet reached a subsequent PBS? Or because photon #1 takes on H polarisation, we write photon #2 as taking on polarisation H as well (even though it hasn't reached a PBS)?